A solution containing.... | JEE Main 2024

JEE-Main_2024

08.04.24_(S1)

Question

A solution containing 10g of an electrolyte $\mathrm{AB}_2$ in 100g of water boils at $100.52^{\circ} \mathrm{C}$. The degree of ionization of the electrolyte (α) is ______× $10^{-1}$ . (nearest integer)
[Given: Molar mass of $\mathrm{AB}_2=200 \mathrm{~g} \mathrm{~mol}^{-1} \cdot \mathrm{~K}_{\mathrm{b}}$ (molal boiling point elevation const. of water) $= 0.52 \mathrm{~K} \mathrm{~kg} \mathrm{~mol}^{-1}$, boiling point of water $=100^{\circ} \mathrm{C} ; \mathrm{AB}_2$ ionises as $\left.\mathrm{AB}_2 \rightarrow \mathrm{~A}^{2+}+2 \mathrm{~B}^{-}\right]$

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Solution

$$ \begin{aligned} & \mathrm{AB}_2 \rightarrow \mathrm{~A}^{+2}+2 \mathrm{~B} \\ & \mathrm{i}=1+(3-1) \alpha \\ & \mathrm{i}=1+2 \alpha \\ & \Delta \mathrm{~T}_{\mathrm{b}}=\mathrm{k}_{\mathrm{b}} \mathrm{im} \\ & 0.52=0.52(1+2 \alpha) \frac{\frac{10}{200}}{\frac{100}{1000}} \\ & 1=(1+2 \alpha) \frac{10}{20} \\ & 2=1+2 \alpha \\ & \alpha=0.5 \end{aligned} $$ Ans. $\alpha=5 \times 10^{-1}$

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JEE Main

Chemistry

Easy

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